In the design of integrated circuits, it is sometimes necessary to use a current source that has a positive temperature coefficient. That is, as the temperature increases, the current level output increases also. One application for such a current source is in the drive for light emitting diodes. Since the brightness of a light emitting diode decreases as the temperature goes up, it is desirable to compensate for this with an increased current as the temperature rises as well.
In FIG. 1, a typical positive temperature coefficient current source is shown. The desired drive current for a light emitting diode is 10mA. To achieve the desired current output in the prior art circuit, the resulting base current needed for transistor N3 if it has a current gain of 100 is 100 microamps. In order for the circuit to operate properly, the current through transistor N1 must be greater than the base current of N3.
It is an object of the present invention to lower the current level required by transistor N1. By lowering the input current level, the power dissipation in the integrated circuit is correspondingly lowered. And yet the positive temperature coefficient characteristics of the current source are maintained.
Referring in greater detail to the circuit of FIG. 1, the output current has a positive temperature coefficient. The voltage generated across the resistor R1 is determined by the difference in base emitter voltages of the transistors N3 and N6. The variation in base emitter voltages is created by the difference in emitter areas between the two transistors. The emitter area of transistor N4 is six times the emitter area in transistor N3. The equation for this circuit is Vbe(N1)+Vbe (N6)+V(R1)=Vbe(N4)+Vbe(N3). Because the collector current of N1 and transistor N4 are nearly equal, they can be subtracted from the equation. This leaves: Vbe(N6)+V(R1)=Vbe(N3). Thus, it is seen that the voltage across resistor R1 is equal to the base emitter voltage of transistor N3 minus the base emitter voltage of transistor N6. Since the collector currents of transistors N3 and N6 are nearly equal, the difference in base emitter voltages can be determined from the ratio of the emitter areas of the two devices. More specifically, Vbe=(kT/Q) ln (A.sub.e N6/A.sub.e N3). Where k is the Boltzmann's constant, T stands for temperature and Q is the charge on an electron. A.sub.e stands for the area of the emitters so that the natural log is being taken of the ratio of the emitter areas of transistors N6 and N3. The output current Iout will be equal to the difference in base emitter voltages of transistors N6 and N3 divided by the resistance of resistor R1. With a typical design using shallow N+ as the resistive material for resistor R1, the numerator, Vbe will rise faster than the denominator, the resistance of R1, as temperature increases. Therefore, the output current will rise as temperature rises.